Deriving Generalized Lomb-Scargle

Standard Lomb-Scargle

In standard Lomb-Scargle, we start with values

$$ \{t,y,dy\} $$

We then compute the mean of the $y$ values using the maximum likelihood:

$$ \mu_y = \frac{\sum_i y_i/dy_i^2}{\sum_i 1/dy_i^2} $$

Then we compute the centered $y$:

$$ \bar{y} \equiv y - \mu_y $$

The model is a simple 1-term sinusoid given by

$$ M(t,\omega~|~\theta) = \theta_0 \sin(\omega t) + \theta_1\cos(\omega t) $$

The likelihood for the dataset is

$$ L(\{t,y,dy\},\omega~|~\theta) = \sum_i \frac{1}{\sqrt{2\pi dy_i^2}} \exp\left[ \frac{-(y_i - M(t_i,\omega~|~\theta)^2}{2dy_i^2} \right] $$

Which leads to the chi-squared function (derived from the log-likelihood)

$$ \chi^2(\omega, \theta) = \sum_i\frac{(\bar{y}_i - M(t_i,\omega~|~\theta)^2}{2dy_i^2} $$

If we re-express the model by defining

$$ X_\omega = \left[ \begin{array}{ll} \sin(\omega t_1) & \cos(\omega t_1)\\ \sin(\omega t_2) & \cos(\omega t_2)\\ \vdots & \vdots \\ \sin(\omega t_N) & \cos(\omega t_N)\\ \end{array} \right],~~~~ y = \left[ \begin{array}{l} y_1 \\ y_2\\ \vdots \\ y_N\\ \end{array} \right] $$

And create the error matrix

$$ \Sigma_y = \left[ \begin{array}{lllll} dy_1^2 & 0 & 0 & \cdots & 0\\ 0 & dy_2^2 & 0 & \cdots & 0\\ 0 & 0 & dy_3^2 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & dy_N^2 \end{array} \right] $$

then our model is given by

$$ M(\omega, \theta) = X_\omega\theta $$

and our $\chi^2$ can be written

$$ \chi^2(\omega, \theta) = (\bar{y} - X_\omega\theta)^T\Sigma_y^{-1}(\bar{y} - X_\omega\theta) $$

Minimizing this cost funciton with respect to $\theta$ gives the maximum likelihood parameters:

$$ \hat{\theta} = (X_\omega^T\Sigma_y^{-1}X_\omega)^{-1}X_\omega^T\Sigma_y^{-1}\bar{y} $$

We can simplify this a bit by defining

$$ X_{\omega,\ast} = \Sigma_y^{-1/2}X_\omega \\ \bar{y}_\ast = \Sigma_y^{-1/2}\bar{y} $$

And the above becomes

$$ \hat{\theta} = (X_{\omega,\ast}^TX_{\omega,\ast})^{-1}X_{\omega,\ast}^T\bar{y}_\ast $$

Now the $\chi^2$ of the model fit is given by

$$ \chi^2(\omega,\hat{\theta}) = \left[ \bar{y}_\ast^T\bar{y}_\ast - \bar{y}_\ast^TX_{\omega,\ast} (X_{\omega,\ast}^TX_{\omega,\ast})^{-1}X_{\omega,\ast}^T\bar{y}_\ast \right] $$

The reference $\chi^2$ is

$$ \chi_0^2 = \bar{y}_\ast^T\bar{y}_\ast $$

So the power $P_{LS} = 1 - \chi^2/\chi_0^2$ is given by

$$ P_{LS}(\omega) = \frac{\bar{y}_\ast^TX_{\omega,\ast} (X_{\omega,\ast}^TX_{\omega,\ast})^{-1}X_{\omega,\ast}^T\bar{y}_\ast}{\bar{y}_\ast^T\bar{y}_\ast} $$

Generalized Lomb-Scargle

The generalized lomb-scargle fits for the mean of $y$ as part of the model, rather than as a separete step.

So what changes is that the $X_\omega$ becomes:

$$ X_\omega = \left[ \begin{array}{lll} 1 & \sin(\omega t_1) & \cos(\omega t_1)\\ 1 & \sin(\omega t_2) & \cos(\omega t_2)\\ \vdots & \vdots & \vdots \\ 1 & \sin(\omega t_N) & \cos(\omega t_N)\\ \end{array} \right] $$

and $\bar{y}$ becomes

$$ \bar{y} = y - X_\omega \mathbf{1}_1 \hat{\theta} $$

where the best-fit parameters $\hat{\theta}$ satisfy

$$ \hat{\theta} = (X_\omega^T\Sigma_y^{-1}X_\omega)^{-1}X_\omega^T\Sigma_y^{-1} y $$

and we have defined,

$$ \mathbf{1}_1 = \left[ \begin{array}{lll} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array} \right] $$

With this change, we write the maximum $\chi^2$ as

$$ \chi^2(\omega) = (y - X_\omega\hat{\theta})^T\Sigma_y^{-1}(y - X_\omega\hat{\theta}) $$

Let's re-express this in terms of $\bar{y}$:

$$ \chi^2(\omega) = (\bar{y} - X_\omega\tilde{\mathbf{1}}_1\hat{\theta})^T\Sigma_y^{-1}(y - X_\omega\tilde{\mathbf{1}}_1\hat{\theta}) $$

Where we have defined

$$ \tilde{\mathbf{1}}_1 = I - \mathbf{1}_1 $$

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